Difference between revisions of "2017 AMC 8 Problems/Problem 24"
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<math>\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152</math> | <math>\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152</math> | ||
| − | ==Solution== | + | ==Solution 1== |
In <math>360</math> days, there are <cmath>360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144</cmath> days without calls. Also note that in the last five days of the year, day <math>361</math> and <math>362</math> also do not have any calls. Thus our answer is <math>144+2 = \boxed{\textbf{(D)}\ 146}</math>. | In <math>360</math> days, there are <cmath>360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144</cmath> days without calls. Also note that in the last five days of the year, day <math>361</math> and <math>362</math> also do not have any calls. Thus our answer is <math>144+2 = \boxed{\textbf{(D)}\ 146}</math>. | ||
~nukelauncher | ~nukelauncher | ||
| + | |||
| + | ==Solution 2== | ||
| + | We use Principle of Inclusion Exclusion. There are <math>365</math> days in the year, and we subtract the days that she gets at least <math>1</math> phone call, which is <math> \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor</math>. | ||
| + | |||
| + | To this result we add the number of days where she gets at least <math>2</math> phone calls in a day because we double subtracted these days. This number is <math>\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor</math>. | ||
| + | |||
| + | We now subtract the number of days where she gets three phone calls, which is <math>\left \lfloor \frac{365}{60} \right \rfloor</math>. | ||
| + | |||
| + | Therefore, our answer is <math>365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{\text{(D) } 146}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 14:30, 22 November 2017
Contents
Problem 24
Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
Solution 1
In 
days, there are ![\[360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144\]](http://latex.artofproblemsolving.com/6/7/8/6783f07de26963cc58034fd375274930f3d2ae9a.png)
days without calls. Also note that in the last five days of the year, day 
and 
also do not have any calls. Thus our answer is 
.
~nukelauncher
Solution 2
We use Principle of Inclusion Exclusion. There are 
days in the year, and we subtract the days that she gets at least 
phone call, which is 
.
To this result we add the number of days where she gets at least 
phone calls in a day because we double subtracted these days. This number is 
.
We now subtract the number of days where she gets three phone calls, which is 
.
Therefore, our answer is 
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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